I've been reading Srednicki's introduction to path integrals and I'm slightly unsure of the notation that he uses for the completeness relation of position eigenstates in the Heisenberg picture. In particular, when he writes $$\mathbf{1}=\int_{-\infty}^{\infty}dq_{j}\lvert q_{j}\rangle\langle q_{j}\rvert$$ is this the instantaneous completeness relation for the position eigenstates in the Heisenberg picture? i.e. Given the instantaneous position eigenstate $\lvert q_{j},t_{j}\rangle =\hat{U}^{\dagger}\lvert q_{j}\rangle$ at time $t_{j}$, then we have $$\int_{-\infty}^{\infty}dq_{j}\lvert q_{j},t_{j}\rangle\langle q_{j},t_{j}\rvert =\int_{-\infty}^{\infty}dq_{j}\hat{U}^{\dagger}\lvert q_{j}\rangle\langle q_{j}\rvert\hat{U}= \hat{U}^{\dagger}\left(\int_{-\infty}^{\infty}dq_{j} \lvert q_{j}\rangle\langle q_{j}\rvert\right)\hat{U} =\hat{U}^{\dagger}\hat{U} =\mathbf{1}$$ where $\lvert q_{j}\rangle$ are the constant position eigenstates corresponding to those in the Schrödinger picture at time $t_{j}$, and thus satisfy the completeness relation $\int_{-\infty}^{\infty}dq_{j}\lvert q_{j}\rangle\langle q_{j}\rvert =\mathbf{1}$. (We've also used that $\hat{U}^{\dagger} \hat{U}=\mathbf{1}$.)

  • $\begingroup$ $\uparrow$ Which page? $\endgroup$ – Qmechanic Aug 31 '15 at 11:01
  • $\begingroup$ @Qmechanic page 57, Chapter 6 ("Path integrals in quantum mechanics") $\endgroup$ – Will Aug 31 '15 at 11:07

I) Let us for clarity use a subscript "$S$" (and "$H$") to denote the Schrödinger (Heisenberg) picture, where bras and kets evolve (are unchanged) and operators are unchanged (evolve), respectively.

Moreover, let us assume that the two pictures coincide at the instant $t_0$ (which Ref. 1 assumes is $t_0=0$).

II) Recall first of all the possibly confusing fact that the Heisenberg instantaneous position eigenstate $|q,t\rangle_H $ does not evolve in time but does depend on the time parameter $t$, such that

$$\hat{Q}(t)_H ~|q,t\rangle_H~=~q~|q,t\rangle_H , \qquad \hat{Q}(t)_H~\equiv~e^{i\hat{H}(t-t_0)/\hbar} \hat{Q}_S ~e^{-i\hat{H}(t-t_0)/\hbar}. $$

The Heisenberg instantaneous position eigenstates satisfy

$$\int_{\mathbb{R}}\! \mathrm{d}q~| q,t\rangle_H ~{}_H\langle q,t|~=~{\bf 1}$$

for all $t$. Moreover

$$|q,t_2\rangle_H~=~e^{i\hat{H}(t_2-t_1)/\hbar} |q,t_1\rangle_H.$$

III) Let us now return to OP's question about whether the ket $|q\rangle $ in Ref. 1 is in the Schrödinger or in the Heisenberg picture? The ket $|q\rangle $ satisfies

$$ \hat{Q}_S~|q\rangle~=~q~|q\rangle. $$

The ket $|q\rangle $ does not evolve in time. Hence it is by definition in the Heisenberg picture. In fact, it is the Heisenberg instantaneous position eigenstate $|q,t_0\rangle_H $. Be aware that $|q\rangle $ often is called a Schrödinger picture position eigenket, cf. Ref. 2, because ... that's what it also is!


  1. M. Srednicki, QFT, 2007; Chapter 6. A prepublication draft PDF file is available here.

  2. R. Dick, Advanced Quantum Mechanics: Materials and Photons, 2012; p.243.

  3. J.J. Sakurai, Modern Quantum Mechanics, 1994; Chapter 2.

  • $\begingroup$ Ah OK, so is it that the instantaneous eigenstates satisfy the completeness relation I gave in my post? I'm just confused by Srednicki's notation of $\lvert q_{j}\rangle$, is this shorthand for $\lvert q_{j}, t_{j}\rangle$? Also, I thought that in the Heisenberg picture the eigenstates of a given basis evolve in time, i.e. $\lvert q_{j},t_{j}\rangle =e^{i\hat{H}t_{j}/\hbar}$, such that the state vector of the system $\lvert\psi\rangle$ remains constant in time? $\endgroup$ – Will Aug 31 '15 at 13:05
  • $\begingroup$ My question arose in particular after reading page 242 in Rainer Dick's book, "Advanced Quantum Mechanics: Materials & Photons: books.google.co.uk/… $\endgroup$ – Will Aug 31 '15 at 13:05
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Aug 31 '15 at 13:42
  • $\begingroup$ Thanks for the updated answer. So is what I put correct (according to what R.Dick has put in his book) about the instantaneous completeness relations? In the Heisenberg picture does $\lvert q_{j},t_{j}\rangle$ simply mean that the system is in a fixed state at the point $q_{j}$ at time $t_{j}$? Is the notation $\lvert q_{j}\rangle$ shorthand for $\lvert q_{j},t_{j}\rangle$? $\endgroup$ – Will Aug 31 '15 at 14:12
  • $\begingroup$ ... Also, in Srednicki's book (chapter 6, page 159) he goes on further to talk about the transition amplitude $\langle q'',t''\lvert \hat{Q}(t_{1})\rvert q',t'\rangle$ introducing a factor of $q(t_{1})$ in the corresponding path integral. By this is it simply meant the eigenvalue $q$ of the position operator $\hat{Q}$ at time $t_{1}$, i.e. $\hat{Q}(t_{1})\lvert q_{1}\rangle = q(t_{1})\lvert q_{1}\rangle$? $\endgroup$ – Will Aug 31 '15 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.